Polar Coordinates and Complex Numbers

Study Sheet

Polar Coordinates and Complex Numbers

Plotting, conversions, polar graphs, trig form, products, and DeMoivre's Theorem

Polar Coordinates

Concept
The Polar Point (r,θ)(r,\theta)
xyAB

A polar point is written (r,θ)(r,\theta), where rr is the directed distance from the origin (the pole) and θ\theta is the angle from the positive xx-axis (the polar axis).

  • [leftmargin=5mm]
  • If r>0r>0, move rr units along the terminal side of θ\theta.
  • If r<0r<0, move ∣r∣|r| units in the opposite direction (add 180∘180^\circ).

Distance and midpoint come from the coordinates.

The point (r,θ)(r,\theta): travel out distance rr at angle θ\theta.

Concept
Many Names for One Point

Every polar point has infinitely many representations:

(r,θ)=(r, θ+360∘k)=(−r, θ+180∘+360∘k),k=0,±1,±2,…(r,\theta) = (r,\ \theta + 360^\circ k) = (-r,\ \theta + 180^\circ + 360^\circ k),\qquad k = 0,\pm1,\pm2,\dots

The pole itself is (0,θ)(0,\theta) for any angle θ\theta.

Example
Example: Other Names for (3,50∘)(3,50^\circ)

Same terminal side, add a revolution: (3, 410∘)(3,\,410^\circ).  Same, go backward: (3, −310∘)(3,\,-310^\circ). Use a negative radius (add 180∘180^\circ): (−3, 230∘)(-3,\,230^\circ).

Concept
Polar ↔\leftrightarrow Rectangular (Points)

With the pole at the origin and polar axis along the positive xx-axis:

 x=rcos⁡θ,y=rsin⁡θ  r=x2+y2,tan⁡θ=yx \boxed{\,x = r\cos\theta,\qquad y = r\sin\theta\,}\qquad\qquad \boxed{\,r = \sqrt{x^2+y^2},\qquad \tan\theta = \dfrac{y}{x}\,}

When finding θ\theta, always check the quadrant of (x,y)(x,y).

Example
Example: Polar →\to Rectangular

Convert (4, 120∘)\left(4,\,120^\circ\right) to rectangular coordinates.

x=4cos⁡120∘=4(−12)=−2,y=4sin⁡120∘=4(32)=23x = 4\cos120^\circ = 4\left(-\tfrac12\right) = -2,\qquad y = 4\sin120^\circ = 4\left(\tfrac{\sqrt3}{2}\right) = 2\sqrt3
⇒(−2, 23)\Rightarrow\quad \mathbf{\left(-2,\ 2\sqrt3\right)}
Example
Example: Rectangular →\to Polar

Convert (−1, 3)\left(-1,\ \sqrt3\right) to polar form (r>0r>0, 0∘≤θ<360∘0^\circ\le\theta<360^\circ).

r=(−1)2+(3)2=4=2r = \sqrt{(-1)^2 + (\sqrt3)^2} = \sqrt{4} = 2

The point is in Quadrant II with reference angle tan⁡−1 ⁣(31)=60∘\tan^{-1}\!\left(\tfrac{\sqrt3}{1}\right)=60^\circ, so θ=180∘−60∘=120∘\theta = 180^\circ - 60^\circ = 120^\circ.

⇒(2, 120∘)\Rightarrow\quad \mathbf{\left(2,\ 120^\circ\right)}
Tip

Tip: A calculator's tan⁡−1\tan^{-1} only returns angles in Quadrants I and IV. Sketch the point first, then adjust θ\theta into the correct quadrant.

Converting Polar and Rectangular Equations

Concept
Equation Conversion Toolkit

Swap between forms with the same relationships, plus these handy substitutions:

r2=x2+y2,x=rcos⁡θ,y=rsin⁡θ,tan⁡θ=yxr^2 = x^2+y^2,\qquad x = r\cos\theta,\qquad y = r\sin\theta,\qquad \tan\theta = \frac{y}{x}

Polar →\to rectangular: multiply by rr or use identities to remove rr and θ\theta. Rectangular →\to polar: replace x,yx,y and simplify; solve for rr if possible.

Example
Example: Polar →\to Rectangular Equation

Convert r=2cos⁡θr = 2\cos\theta. Multiply both sides by rr:

r2=2rcos⁡θ  ⟹  x2+y2=2x  ⟹  x2−2x+y2=0  ⟹  (x−1)2+y2=1r^2 = 2r\cos\theta \;\Longrightarrow\; x^2+y^2 = 2x \;\Longrightarrow\; x^2-2x+y^2 = 0 \;\Longrightarrow\; \mathbf{(x-1)^2 + y^2 = 1}

A circle of radius 11 centered at (1,0)(1,0).

Example
Example: Rectangular →\to Polar Equation

Convert x2+y2=9x^2 + y^2 = 9 and the line x=3x = 3.

x2+y2=9  ⟹  r2=9  ⟹  r=3x^2+y^2 = 9 \;\Longrightarrow\; r^2 = 9 \;\Longrightarrow\; \mathbf{r = 3}
x=3  ⟹  rcos⁡θ=3  ⟹  r=3sec⁡θx = 3 \;\Longrightarrow\; r\cos\theta = 3 \;\Longrightarrow\; \mathbf{r = 3\sec\theta}
Tip

Remember: r=ar = a is a circle of radius aa; θ=α\theta = \alpha is a line through the pole. Vertical/horizontal lines become r=asec⁡θr = a\sec\theta or r=acsc⁡θr = a\csc\theta.

Graphs of Polar Equations

Concept
Common Polar Curves
  • [leftmargin=5mm]
  • Circles: r=ar=a (centered at pole); r=acos⁡θr=a\cos\theta or r=asin⁡θr=a\sin\theta (through the pole, diameter ∣a∣|a|).
  • Cardioids: r=a±acos⁡θr = a \pm a\cos\theta or r=a±asin⁡θr = a\pm a\sin\theta --- a heart shape with one dimple at the pole.
  • Limaçons: r=a±bcos⁡θr = a \pm b\cos\theta or r=a±bsin⁡θr = a\pm b\sin\theta. Compare ab\tfrac{a}{b}: ab<1\tfrac{a}{b}<1 inner loop; ab=1\tfrac{a}{b}=1 cardioid; 1<ab<21<\tfrac{a}{b}<2 dimpled; ab≥2\tfrac{a}{b}\ge 2 convex.
  • Roses: r=acos⁡(nθ)r = a\cos(n\theta) or r=asin⁡(nθ)r = a\sin(n\theta). Petal count: nn petals if nn is odd, 2n2n petals if nn is even.

Cardioid r=1+cos⁡θr = 1+\cos\theta: maximum r=2r=2 at θ=0∘\theta=0^\circ, dimple at the pole when θ=180∘\theta=180^\circ.

Rose r=2cos⁡2θr = 2\cos 2\theta: since n=2n=2 is even, it has 2n=42n = \mathbf{4} petals, each of length 22.

Tip

Tip: To sketch, make a quick table of θ=0∘,30∘,45∘,60∘,90∘,…\theta = 0^\circ,30^\circ,45^\circ,60^\circ,90^\circ,\dots and plot (r,θ)(r,\theta). Note where r=0r=0 (curve passes through the pole) and where ∣r∣|r| is largest.

Complex Numbers and Trigonometric Form

Concept
The Complex Plane

A complex number z=a+biz = a + bi is plotted as the point (a,b)(a,b): the real part aa on the horizontal axis, the imaginary part bb on the vertical axis. Its

modulus ∣z∣=r=a2+b2,argument θ with tan⁡θ=ba  (choose by quadrant).\text{\textbf{modulus} } |z| = r = \sqrt{a^2+b^2},\qquad \text{\textbf{argument} } \theta \text{ with } \tan\theta = \frac{b}{a}\ \ (\text{choose by quadrant}).

z=a+biz = a+bi has modulus rr (distance to origin) and argument θ\theta.

Concept
Trigonometric (Polar) Form
z=a+bi=r(cos⁡θ+isin⁡θ),where a=rcos⁡θ,  b=rsin⁡θ.z = a + bi = r(\cos\theta + i\sin\theta),\qquad \text{where } a = r\cos\theta,\ \ b = r\sin\theta.

The shorthand r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta) is often written r cis θr\,\text{cis}\,\theta.

Example
Example: Rectangular →\to Trig Form

Write z=−1+i3z = -1 + i\sqrt3 in trigonometric form.

r=(−1)2+(3)2=2.r = \sqrt{(-1)^2 + (\sqrt3)^2} = 2.

The point (−1,3)(-1,\sqrt3) is in Quadrant II with reference angle 60∘60^\circ, so θ=120∘\theta = 120^\circ.

⇒z=2(cos⁡120∘+isin⁡120∘)\Rightarrow\quad z = \mathbf{2\left(\cos120^\circ + i\sin120^\circ\right)}
Tip

Tip: Going from trig form back to a+bia+bi, just evaluate: z=rcos⁡θ+(rsin⁡θ) iz = r\cos\theta + (r\sin\theta)\,i.

Products and Quotients in Trig Form

Concept
Multiply and Divide by Modulus and Angle

For z1=r1(cos⁡θ1+isin⁡θ1)z_1 = r_1(\cos\theta_1 + i\sin\theta_1) and z2=r2(cos⁡θ2+isin⁡θ2)z_2 = r_2(\cos\theta_2 + i\sin\theta_2):

z1z2=r1r2[cos⁡(θ1+θ2)+isin⁡(θ1+θ2)]z_1 z_2 = r_1 r_2\big[\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)\big]
z1z2=r1r2[cos⁡(θ1−θ2)+isin⁡(θ1−θ2)]\frac{z_1}{z_2} = \frac{r_1}{r_2}\big[\cos(\theta_1-\theta_2) + i\sin(\theta_1-\theta_2)\big]

Multiply the moduli, add the angles (divide moduli, subtract angles).

Example
Example: Product

Let z1=3(cos⁡40∘+isin⁡40∘)z_1 = 3(\cos40^\circ + i\sin40^\circ) and z2=2(cos⁡80∘+isin⁡80∘)z_2 = 2(\cos80^\circ + i\sin80^\circ).

z1z2=(3)(2)[cos⁡(40∘+80∘)+isin⁡(40∘+80∘)]=6(cos⁡120∘+isin⁡120∘)z_1 z_2 = (3)(2)\big[\cos(40^\circ+80^\circ) + i\sin(40^\circ+80^\circ)\big] = \mathbf{6\left(\cos120^\circ + i\sin120^\circ\right)}

In a+bia+bi: 6(−12+i32)=−3+3i36\left(-\tfrac12 + i\tfrac{\sqrt3}{2}\right) = -3 + 3i\sqrt3.

Tip

Remember: If a subtracted angle comes out negative or over 360∘360^\circ, add or subtract 360∘360^\circ to land in [0∘,360∘)[0^\circ,360^\circ).

DeMoivre's Theorem: Powers and Roots

Concept
DeMoivre's Theorem (Powers)

For z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta) and any positive integer nn:

 zn=rn(cos⁡nθ+isin⁡nθ) \boxed{\,z^n = r^n\big(\cos n\theta + i\sin n\theta\big)\,}

Raise the modulus to the power; multiply the angle by nn.

Example
Example: Power

Compute (1+i3)4\left(1 + i\sqrt3\right)^4. First r=1+3=2r = \sqrt{1+3} = 2 and θ=60∘\theta = 60^\circ (Quadrant I).

z4=24(cos⁡(4⋅60∘)+isin⁡(4⋅60∘))=16(cos⁡240∘+isin⁡240∘)z^4 = 2^4\big(\cos(4\cdot 60^\circ) + i\sin(4\cdot 60^\circ)\big) = 16\big(\cos240^\circ + i\sin240^\circ\big)
=16(−12−i32)=−8−8i3= 16\left(-\tfrac12 - i\tfrac{\sqrt3}{2}\right) = \mathbf{-8 - 8i\sqrt3}
Concept
The nn Distinct nnth Roots

Every nonzero z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta) has exactly nn complex nnth roots:

wk=rn[cos⁡ ⁣(θ+360∘kn)+isin⁡ ⁣(θ+360∘kn)],k=0,1,2,…,n−1.w_k = \sqrt[n]{r}\left[\cos\!\left(\frac{\theta + 360^\circ k}{n}\right) + i\sin\!\left(\frac{\theta + 360^\circ k}{n}\right)\right],\qquad k = 0,1,2,\dots,n-1.

All roots share modulus rn\sqrt[n]{r} and are spaced 360∘n\dfrac{360^\circ}{n} apart around a circle.

Example
Example: All Cube Roots of 8i8i

Write 8i=8(cos⁡90∘+isin⁡90∘)8i = 8(\cos90^\circ + i\sin90^\circ), so r=8r=8 and 83=2\sqrt[3]{8} = 2. Angles: 90∘+360∘k3\dfrac{90^\circ + 360^\circ k}{3} for k=0,1,2k=0,1,2, i.e. 30∘, 150∘, 270∘30^\circ,\,150^\circ,\,270^\circ.

w0=2(cos⁡30∘+isin⁡30∘)=3+iw_0 = 2(\cos30^\circ + i\sin30^\circ) = \sqrt3 + i
w1=2(cos⁡150∘+isin⁡150∘)=−3+iw_1 = 2(\cos150^\circ + i\sin150^\circ) = -\sqrt3 + i
w2=2(cos⁡270∘+isin⁡270∘)=−2iw_2 = 2(\cos270^\circ + i\sin270^\circ) = -2i

The three roots: 3+i,  −3+i,  −2i\mathbf{\sqrt3 + i,\ \ -\sqrt3 + i,\ \ -2i}.

Tip

Tip: Find the first root (k=0k=0), then just add 360∘n\dfrac{360^\circ}{n} to the angle repeatedly to get the rest --- the modulus never changes.

Going Deeper: Advanced Polar & Complex Ideas

Concept
Roots of Unity

The nnth roots of unity are the nn solutions of zn=1z^n = 1. Writing ω=cos⁡ ⁣360∘n+isin⁡ ⁣360∘n\omega = \cos\!\frac{360^\circ}{n} + i\sin\!\frac{360^\circ}{n} (the primitive root), every root is a power of ω\omega:

ωk=cos⁡ ⁣(360∘kn)+isin⁡ ⁣(360∘kn),k=0,1,2,…,n−1.\omega^k = \cos\!\left(\frac{360^\circ k}{n}\right) + i\sin\!\left(\frac{360^\circ k}{n}\right),\qquad k = 0,1,2,\dots,n-1.

They sit at the vertices of a regular nn-gon on the unit circle, starting at ω0=1\omega^0 = 1. Two beautiful facts (for n≥2n\ge 2):

 ∑k=0n−1ωk=0  ∏k=1n−1(1−ωk)=n \boxed{\ \sum_{k=0}^{n-1}\omega^k = 0\ }\qquad\qquad \boxed{\ \prod_{k=1}^{n-1}\bigl(1-\omega^k\bigr) = n\ }

The five 55th roots of unity ω0,…,ω4\omega^0,\dots,\omega^4 form a regular pentagon on the unit circle; their sum is the center, 00.

Example
Example: Sum and Product of the 55th Roots of Unity

Here ω=cos⁡72∘+isin⁡72∘\omega = \cos72^\circ + i\sin72^\circ, and the five roots are

ω0, ω1, ω2, ω3, ω4  =  cis 0∘, cis 72∘, cis 144∘, cis 216∘, cis 288∘.\omega^0,\ \omega^1,\ \omega^2,\ \omega^3,\ \omega^4 \;=\; \text{cis}\,0^\circ,\ \text{cis}\,72^\circ,\ \text{cis}\,144^\circ,\ \text{cis}\,216^\circ,\ \text{cis}\,288^\circ.

Sum. Because z5−1=(z−1)(z4+z3+z2+z+1)z^5-1 = (z-1)(z^4+z^3+z^2+z+1), the roots satisfy z4+z3+z2+z+1=0z^4+z^3+z^2+z+1=0; comparing to ∑ωk\sum\omega^k, the coefficient of z4z^4 is 11, so

ω0+ω1+ω2+ω3+ω4=0.\omega^0+\omega^1+\omega^2+\omega^3+\omega^4 = 0.

Geometrically the five equally spaced vectors cancel by symmetry. Product. Factor z5−1=(z−1)∏k=14(z−ωk)z^5-1 = (z-1)\prod_{k=1}^{4}(z-\omega^k), so ∏k=14(z−ωk)=z4+z3+z2+z+1\prod_{k=1}^{4}(z-\omega^k) = z^4+z^3+z^2+z+1. Setting z=1z=1:

∏k=14(1−ωk)=1+1+1+1+1=5.\prod_{k=1}^{4}\bigl(1-\omega^k\bigr) = 1+1+1+1+1 = \mathbf{5}.
Concept
The Identity zn+z−n=2cos⁡nθz^n + z^{-n} = 2\cos n\theta

If z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta lies on the unit circle, then 1z=zˉ=cos⁡θ−isin⁡θ\tfrac1z = \bar z = \cos\theta - i\sin\theta. By DeMoivre,

zn=cos⁡nθ+isin⁡nθ,z−n=cos⁡nθ−isin⁡nθ,z^n = \cos n\theta + i\sin n\theta,\qquad z^{-n} = \cos n\theta - i\sin n\theta,

so adding and subtracting gives the two workhorse identities

 zn+z−n=2cos⁡nθ  zn−z−n=2isin⁡nθ \boxed{\,z^n + z^{-n} = 2\cos n\theta\,}\qquad\qquad \boxed{\,z^n - z^{-n} = 2i\sin n\theta\,}

These turn powers of cos⁡θ\cos\theta and sin⁡θ\sin\theta into sums of cos⁡nθ\cos n\theta and sin⁡nθ\sin n\theta (and back).

Example
Example: Using DeMoivre to Derive cos⁡3θ\cos 3\theta and sin⁡3θ\sin 3\theta

Expand (cos⁡θ+isin⁡θ)3(\cos\theta + i\sin\theta)^3 two ways. By DeMoivre, it equals cos⁡3θ+isin⁡3θ\cos 3\theta + i\sin 3\theta. By the binomial theorem (using i2=−1i^2=-1, i3=−ii^3=-i):

(cos⁡θ+isin⁡θ)3=cos⁡3θ+3icos⁡2θsin⁡θ−3cos⁡θsin⁡2θ−isin⁡3θ.(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta.

Match real parts and imaginary parts:

cos⁡3θ=cos⁡3θ−3cos⁡θsin⁡2θ=4cos⁡3θ−3cos⁡θ,\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = \mathbf{4\cos^3\theta - 3\cos\theta},
sin⁡3θ=3cos⁡2θsin⁡θ−sin⁡3θ=3sin⁡θ−4sin⁡3θ,\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta = \mathbf{3\sin\theta - 4\sin^3\theta},

where the final forms use sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1-\cos^2\theta and cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1-\sin^2\theta.

Reminder — The binomial theorem:  (x+y)n=∑k=0n(nk)x n−kyk(x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{\,n-k}y^{k}
Concept
Apollonius Circles: the Locus ∣z−a∣=k ∣z−b∣|z-a| = k\,|z-b|

The set of points zz whose distance to aa is a fixed multiple k>0k>0 of its distance to bb is:

  • [leftmargin=5mm]
  • the perpendicular bisector of aa and bb when k=1k=1 (equal distances);
  • an Apollonius circle when k≠1k\ne 1 --- squaring ∣z−a∣2=k2∣z−b∣2|z-a|^2 = k^2|z-b|^2 and expanding z=x+iyz=x+iy produces x2+y2+(⋯ )x+(⋯ )y+(⋯ )=0x^2+y^2+(\cdots)x+(\cdots)y+(\cdots)=0, the equation of a circle.

The circle is symmetric about the line through aa and bb and shrinks toward the nearer point as k→0k\to0 or k→∞k\to\infty.

Example
Example: Describe the Locus ∣z−1∣=2 ∣z+1∣|z-1| = 2\,|z+1|

Let z=x+iyz = x+iy, with a=1a=1 and b=−1b=-1. Square both sides:

(x−1)2+y2=4[(x+1)2+y2].(x-1)^2 + y^2 = 4\bigl[(x+1)^2 + y^2\bigr].

Expand and collect:

x2−2x+1+y2=4x2+8x+4+4y2  ⟹  3x2+3y2+10x+3=0.x^2-2x+1+y^2 = 4x^2+8x+4+4y^2 \;\Longrightarrow\; 3x^2 + 3y^2 + 10x + 3 = 0.

Divide by 33 and complete the square:

x2+103x+y2+1=0  ⟹  (x+53)2+y2=259−1=169.x^2 + \tfrac{10}{3}x + y^2 + 1 = 0 \;\Longrightarrow\; \left(x+\tfrac53\right)^2 + y^2 = \tfrac{25}{9}-1 = \tfrac{16}{9}.

So the locus is a circle centered at (−53, 0)\left(-\tfrac53,\,0\right) with radius 43\tfrac43.

Concept
Area Enclosed by a Polar Curve

The region swept by the ray to a polar curve r=f(θ)r = f(\theta) as θ\theta runs from α\alpha to β\beta has area

 A=12∫αβr2 dθ=12∫αβ[f(θ)]2 dθ \boxed{\ A = \frac12\int_{\alpha}^{\beta} r^2\,d\theta = \frac12\int_{\alpha}^{\beta} \bigl[f(\theta)\bigr]^2\,d\theta\ }

This comes from summing thin circular sectors of angle dθd\theta and radius rr, each of area 12r2 dθ\tfrac12 r^2\,d\theta. Setup cautions: choose α,β\alpha,\beta that trace the region exactly once; for one petal of a rose, integrate between consecutive zeros of rr; use symmetry to integrate a half and double.

Example
Example: Set Up the Area of One Petal of r=2cos⁡2θr = 2\cos 2\theta

One petal is traced as rr goes from 00 up to its max and back to 00. Solve r=0r=0: cos⁡2θ=0\cos 2\theta = 0 at 2θ=±90∘2\theta = \pm 90^\circ, i.e. θ=−45∘\theta = -45^\circ and θ=45∘\theta = 45^\circ (the petal centered on the polar axis). Thus

Aone petal=12∫−π/4π/4(2cos⁡2θ)2 dθ=12∫−π/4π/44cos⁡22θ dθ.A_{\text{one petal}} = \frac12\int_{-\pi/4}^{\pi/4} \bigl(2\cos 2\theta\bigr)^2\,d\theta = \frac12\int_{-\pi/4}^{\pi/4} 4\cos^2 2\theta\,d\theta.

By symmetry this equals 2∫0π/4cos⁡22θ dθ\displaystyle 2\int_{0}^{\pi/4}\cos^2 2\theta\,d\theta, the required set-up. (Evaluating with cos⁡22θ=12(1+cos⁡4θ)\cos^2 2\theta = \tfrac12(1+\cos4\theta) gives A=π4A = \tfrac{\pi}{4}.)

Concept
Intersections of Polar Curves

To find where r=f(θ)r = f(\theta) and r=g(θ)r = g(\theta) meet, solving f(θ)=g(θ)f(\theta) = g(\theta) is not enough. Because a single point has many polar names, also check:

  • [leftmargin=5mm]
  • the pole separately --- it lies on a curve if r=0r=0 for some θ\theta, even if the θ\theta values differ;
  • alternate representations, e.g. f(θ)=−g(θ+180∘)f(\theta) = -g(\theta + 180^\circ), and shifts by full turns.

Always sketch both curves to confirm the true intersection points.

Example
Example: Intersections of r=1+cos⁡θr = 1 + \cos\theta and r=3cos⁡θr = 3\cos\theta

Set equal: 1+cos⁡θ=3cos⁡θ⇒1=2cos⁡θ⇒cos⁡θ=121+\cos\theta = 3\cos\theta \Rightarrow 1 = 2\cos\theta \Rightarrow \cos\theta = \tfrac12, so θ=60∘, 300∘\theta = 60^\circ,\,300^\circ. Then r=3cos⁡60∘=32r = 3\cos60^\circ = \tfrac32, giving points (32,60∘)\left(\tfrac32,60^\circ\right) and (32,300∘)\left(\tfrac32,300^\circ\right). Check the pole. On the cardioid, r=0r=0 when cos⁡θ=−1\cos\theta = -1 (θ=180∘\theta=180^\circ); on the circle, r=0r=0 when cos⁡θ=0\cos\theta=0 (θ=90∘\theta=90^\circ). Different θ\theta, but both curves pass through the pole, so the pole is a third intersection point that the algebra missed.

Tip

Sum of the roots of zn=wz^n = w. All nn roots are w0, w0ω, w0ω2,…,w0ωn−1w_0,\ w_0\omega,\ w_0\omega^2,\dots,w_0\omega^{n-1}, where w0w_0 is one root and ω\omega is a primitive nnth root of unity. Factoring out w0w_0,

∑k=0n−1wk=w0∑k=0n−1ωk=w0⋅0=0(n≥2).\sum_{k=0}^{n-1} w_k = w_0\sum_{k=0}^{n-1}\omega^k = w_0\cdot 0 = \mathbf{0}\qquad(n\ge 2).

Equivalently, zn−w=zn+0⋅zn−1+⋯z^n - w = z^n + 0\cdot z^{n-1}+\cdots has no zn−1z^{n-1} term, so by Vieta the roots sum to 00. Their product is (−1)n+1w(-1)^{n+1}w.

Reminder — Vieta's formulas:  r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}

Formulas, Proofs & Tips

Tip
Polar form and De Moivre’s theorem
z=r(cos⁡θ+isin⁡θ),zn=rn(cos⁡nθ+isin⁡nθ)z=r(\cos\theta+i\sin\theta),\qquad z^{n}=r^{n}\big(\cos n\theta+i\sin n\theta\big)

What it means. A complex number is a length and a direction; powering it powers the length and multiplies the angle.

Example. (1+i)4(1+i)^4: here r=2r=\sqrt2, θ=45∘\theta=45^\circ, so r4=4r^4=4 and 4θ=180∘4\theta=180^\circ, giving −4-4.

Why it works. Multiplying two complex numbers in polar form and applying the sine and cosine sum formulas produces r1r2r_1r_2 with angle θ1+θ2\theta_1+\theta_2 — multiplication adds angles. Repeating nn times gives De Moivre.

Tip. r=x2+y2r=\sqrt{x^2+y^2} and θ=arctan⁡yx\theta=\arctan\tfrac{y}{x} — but check the quadrant, since arctan⁡\arctan only returns two of them.